Help with Thermochemistry Problems: Cu & Water

[tandoorichicken]

I don't remember at all how to do these thermochemistry problems. Can someone please guide me through?

You add 30.0 g of Cu (Specific Heat Capacity = 0.3846 J/K·g) at 92.0°C to 150.0 mL of water at 25.0°C in a very insulating calorimeter. Assuming the calorimeter constant (Cc) is zero, and knowing the Specific Heat Capacity of water is 4.18 J/K·g, what was the final temperature of the combined water/Cu metal solution?

 

[learningphysics]

I don't remember at all how to do these thermochemistry problems. Can someone please guide me through?

You add 30.0 g of Cu (Specific Heat Capacity = 0.3846 J/K·g) at 92.0°C to 150.0 mL of water at 25.0°C in a very insulating calorimeter. Assuming the calorimeter constant (Cc) is zero, and knowing the Specific Heat Capacity of water is 4.18 J/K·g, what was the final temperature of the combined water/Cu metal solution?

The heat change in a particular substance (the heat entering the substance) is mc(Tf-Ti), where Tf is the final temperature of that substance, and Ti is in the initial temperature of that substance, and c is the specific heat capacity of that substance.

The total heat change of any group of isolated substances adds to 0.

So (m1)(c1)(Tf1-Ti1)+(m2)(c2)(Tf2-Ti2)+...=0

Hope this helps, and you can apply it to your particular problem.

 

[xanthym]

I don't remember at all how to do these thermochemistry problems. Can someone please guide me through?

You add 30.0 g of Cu (Specific Heat Capacity = 0.3846 J/K·g) at 92.0°C to 150.0 mL of water at 25.0°C in a very insulating calorimeter. Assuming the calorimeter constant (Cc) is zero, and knowing the Specific Heat Capacity of water is 4.18 J/K·g, what was the final temperature of the combined water/Cu metal solution?

Let T_f be Final Temperature of combined water/metal solution. Then from conservation of energy:
{Heat Lost by Metal} = {Heat Gained by Water}
::: ⇒ {Metal Mass}*{Metal Heat Capacity}*{(92 + 273) - T_f} =
= {Water Mass}*{Water Heat Capacity}*{T_f - (25 + 273)}
::: ⇒ {30.0 g}*{0.3846 J/(K*g)}*{(92 + 273) - T_f} = {150.0 g}*{4.18 J/(K*g)}*{T_f - (25 + 273)}
::: ⇒ {11.54}*{365 - T_f} = {627}*{T_f - 298}
::: ⇒ (4212) - (11.54)*T_f = (627)*T_f - (186,800)
::: ⇒ (191,000) = (638.5)*T_f
::: ⇒ T_f = (299.1 degK) = (26.1 degC)


~~